Why Donor Reuse and Ties Matter
Predictive mean matching (PMM) selects an observed case whose prediction score is close to the score for a case with a missing outcome, then uses that observed case’s outcome as the imputation. The candidate pool, matching distance, and donor draw therefore determine which observed values can be copied. When scores tie or only a few donors are eligible, the details of this selection matter: different tie rules can produce different pools, and repeated selections can make some observed outcomes appear many times in the completed data.
The focus here is the selection step after the PMM scores and a pool size have been set. The pool size is denoted by \(d\); the \(d\) closest eligible observed cases are intended to be candidates for a missing case. How to choose \(d\) is a separate decision. A tie rule must clarify what happens when more than \(d\) cases are equally close at the pool boundary.
Assume the observed donor cases are eligible under the imputation model, their scores and distances have been calculated consistently, and the specified candidate-pool size \(d\) does not exceed the number of eligible observed cases. A random donor draw from a pool is ordinarily understood to give each candidate equal probability, unless the method explicitly specifies otherwise. These conditions describe the selection calculation; they do not guarantee that the imputation model or its predictors are adequate.
Resolving Equal Distances
If two candidates have equal distances but both fall within the \(d\) closest, the tie does not change pool membership. The donor is still drawn from the resulting candidate pool according to the stated sampling rule. A tie matters especially at the boundary: for example, if the \(d\)th and \((d+1)\)th cases have the same distance, ranking alone does not identify a unique set of exactly \(d\) cases.
There are three common ways to resolve a boundary tie. A procedure can randomly select enough of the tied cases to keep the pool size at \(d\); it can include every case tied at the boundary, making the pool larger than \(d\); or it can use a fixed rule, such as case order. The first two can be defensible when specified in advance. A fixed rule is reproducible but may systematically favor cases for reasons unrelated to their scores. Quietly allowing software order or row order to decide is not a statistical tie policy.
A tie can also result from rounded scores or distances. If distances are rounded for display, equal printed values need not be exactly equal internally. Conversely, software may define a numerical tolerance and treat sufficiently close distances as tied. The analyst should distinguish the displayed precision from the actual comparison rule and document any tolerance that changes pool membership.
Worked Example: A Tie at the Candidate-Pool Boundary
Suppose a missing case has four eligible observed candidates, and the matching distances are \(0.10\), \(0.20\), \(0.20\), and \(0.35\). The planned pool size is \(d=2\). The two closest distances are \(0.10\) and \(0.20\), but two cases share the boundary distance \(0.20\).
If the method requires exactly two candidates and randomly breaks the tie, the closest case is always included, and one of the two cases at distance \(0.20\) is selected with probability \(1/2\). For example, if their observed outcomes are 12 and 18, each tied case has probability \(1/2\) of entering the pool. The subsequent donor draw is made from the two selected candidates, not automatically from all three.
If the policy includes all boundary ties, the pool contains three cases: the closest case and both cases at distance \(0.20\). With an equal-probability draw from that pool, each candidate has probability \(1/3\) of being selected as donor. This differs from random tie-breaking followed by a draw: in that procedure, the closest case has probability \(1/2\) of selection, while each tied case has probability \((1/2)(1/2)=1/4\).
Thus, “break ties randomly” and “include all tied cases” do not generally produce the same donor probabilities. The tie rule needs to be explicit, especially when the tied cases have different observed outcomes.
With and Without Replacement
For each missing case, a donor is selected from that case’s candidate pool. Under sampling with replacement across recipients, a donor selected for one missing case remains available for later missing cases. The same observed case may therefore be used repeatedly. Independent random draws for different recipients are a common implementation: “with replacement” does not mean that the donor outcome is altered; it means the case is not removed from later candidate pools.
Under sampling without replacement across recipients, a selected donor is removed from later selection, at least within whatever group or set of recipients the rule specifies. This can limit repeated donor use, but it changes the selection probabilities and can be infeasible if more recipients need donors than there are available cases. Because recipients may have different candidate pools, a global without-replacement rule can also make later recipients’ available choices depend on earlier draws. It is not interchangeable with independent PMM draws.
A without-replacement rule might instead be applied only within a particular batch, or only among recipients who share a candidate pool. Those are distinct procedures. Reporting merely “without replacement” is incomplete unless the scope is clear. For ordinary independent PMM donor draws, reuse across missing cases is allowed; a donor can be drawn again even if it has already supplied an imputed value.
What a Small Pool Implies
Repeated use is more likely when many recipients draw from a small common pool. A simple calculation makes this precise. Suppose there are \(m\) candidate donors, and \(q\) recipients each independently select one donor uniformly with replacement. Let \(N_j\) be the number of times donor \(j\) is selected. For any one donor, each draw selects it with probability \(1/m\), so \(N_j\) has a binomial distribution with \(q\) trials and success probability \(1/m\). In particular, its expected use count is \(q/m\).
The first expression is an average over repeated runs of the random selection procedure, not a promise that donors will be used equally in one completed data set. The second expression gives the probability that a specified donor is never selected. Hence the probability that this donor is selected at least once is \(1-(1-1/m)^q\). When the observed outcomes differ, repeated selection can also give particular outcome values more representation among imputations than their number of distinct source cases would suggest.
Worked Example: Expected Reuse in a Small Pool
Consider six missing cases, each with the same three eligible donors, A, B, and C. Assume independent, equally likely donor draws with replacement. For any specified donor, say A, \(m=3\) and \(q=6\).
The expected number of times A is selected is \(E(N_A)=6(1/3)=2\) selections. Its probability of never being selected is \((1-1/3)^6=(2/3)^6=64/729\approx0.08779\). Therefore, the probability that A is selected at least once is \(1-64/729=665/729\approx0.9122\).
The chance that A is selected at least twice is \[ 1-P(N_A=0)-P(N_A=1) =1-\left(\frac{2}{3}\right)^6 -6\left(\frac{1}{3}\right)\left(\frac{2}{3}\right)^5 =\frac{473}{729}\approx0.6488. \] So there is about a 64.88% probability that this particular donor supplies at least two of the six imputations. By symmetry, the same probabilities apply separately to B and C. These events are not independent, and the calculation does not say that every donor will be reused in any particular run.
If an analyst tried to assign six recipients to these three donors without replacement, the rule would fail after at most three assignments. This is not a small technical adjustment to the same procedure: it is an incompatible assignment constraint unless the rule is changed, for example by defining separate groups or allowing reuse.
Conditional on fixed candidate pools and independent draws, repeated donor selection is a consequence of the sampling rule, not evidence of a coding error by itself. It can nevertheless make the imputed values less varied than a user expects, particularly when the eligible cases have few distinct outcomes. Repeated values should be interpreted as repeated use of observed donor outcomes, not as new independent observations from the original population.
Checking the Selection Process
Useful checks focus on the actual donor choices, not just on whether the imputation routine completed. For each imputation cycle or completed data set, inspect how often each observed case supplied a donor, and whether the same outcome values were consequently copied many times. Also check the number of distinct eligible donors for each recipient, the distribution of distances, and whether boundary ties occur often. When a tie rule is random, repeated runs with the same model and data can help reveal how much the rule changes donor assignments.
- Confirm that every selected donor belongs to the intended candidate pool for that recipient.
- Tabulate donor-use counts and identify whether a few cases account for a large share of imputations.
- Review exact or tolerance-based ties at the pool boundary and verify that the documented rule was applied.
- Check whether limited eligibility is concentrated in particular predictor profiles or groups.
- Compare results under plausible, explicitly defined tie policies when ties materially affect donor selection.
A high reuse count is not automatically a failure. It may be expected when \(q\) is large relative to \(m\), or when different recipients have overlapping candidate pools. It is a warning to investigate if one donor dominates unexpectedly, if donor eligibility is being misapplied, or if ties and row ordering produce unintended systematic selection. These checks describe donor selection; they do not replace assessment of the imputation model and its assumptions.
Alternatives, Limitations, and a Common Mistake
Randomly resolving a boundary tie while keeping exactly \(d\) candidates preserves the intended pool size but introduces randomness in who enters it. Including all boundary ties avoids arbitrarily excluding equally close cases, but makes the pool size variable and can alter the relative probability of selection. A fixed tie-break is easy to reproduce but can be sensitive to data ordering. Each option has a different implication; choose and report one rather than treating them as equivalent.
Independent draws with replacement are simple and allow each recipient’s draw to follow its own candidate pool, but they permit repeated use. A no-reuse assignment can reduce duplicate donor selections where feasible, but it links recipients’ selections and may leave a recipient with no available donor. Neither policy repairs a pool that is too small or a prediction score that ranks unsuitable cases. If eligibility is sparse, the selection rule alone cannot create information from cases that are not observed.
Common mistake: “The same donor was used three times, so the PMM algorithm must have sampled without replacement incorrectly.” This reverses the logic. Repeated use is allowed under sampling with replacement and is often expected from a small pool. The correction is to check the specified replacement rule, candidate pools, and draw records. Treat repetition as a diagnostic feature to interpret, not as proof of a bug or proof that the imputation is valid.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions below.
- A pool is intended to contain two donors, but three cases tie for the second-smallest distance. Name two defensible tie policies and explain why their selection probabilities can differ.
- Four recipients independently draw with replacement from a common pool of two donors. What is the expected number of times a specified donor is used, and what is its probability of being unused?
- Why can a global without-replacement policy be infeasible for six recipients and three donors?
- Does observing repeated use of one donor prove that PMM was implemented incorrectly? State what should be checked instead.
Solutions. (1) Randomly select enough of the tied cases to keep exactly two candidates, or include all boundary-tied cases. Including all ties changes the pool size and the equal-probability donor draw; random tie-breaking gives only a subset of tied cases a chance to enter the pool on each resolution, so the resulting donor probabilities can differ. (2) Here \(q=4\) and \(m=2\), so the expected count is \(q/m=2\). The probability of no use is \((1-1/2)^4=1/16=0.06250\). (3) Without replacement, each donor can be assigned at most once, so three donors cannot supply six assignments. (4) No. Check whether reuse is allowed, verify the recipient-specific pools and draw records, and examine pool sizes, ties, and donor-use counts. Reuse is expected under with-replacement draws and is not alone evidence of an error.